Answered

B. A radius vector of a point A relative to the origin varies with time t as r = ati — bt2 j, where a and b are positive constants, and i and j are the unit vectors of the x and y axes.

Find:
(a) the equation of the point's trajectory y (x); plot this function;
(b) the time dependence of the velocity v and acceleration w vectors,
as well as of the moduli of these quantities;
(c) the time dependence of the angle a between the vectors w and v;
(d) the mean velocity vector averaged over the first t seconds of
motion, and the modulus of this vector​

Sagot :

MasterOfMathProblemviz

✒️RADIUS

It is given that

  • [tex]\tt :\longmapsto\:\vec{r} = a t \: \hat{i} - b {t}^{2} \: \hat{j}[/tex]

So,

On comparing with,

  • [tex]\tt :\longmapsto\:\vec{r} = x \: \hat{i} + y \: \hat{j}[/tex]

we get

  • [tex]\tt :\implies\:x = at \: \: and \: \: y = - b {t}^{2} [/tex]

Now, eliminate 't' from these two conditions,

  • [tex]\tt :\longmapsto\: \: as \: x = at \: \tt :\implies\:t = \dfrac{x}{a} [/tex]

So,

  • [tex]\tt :\implies\:y = - b \times \dfrac{ {x}^{2} }{ {a}^{2} } [/tex]

  • [tex]\tt :\implies\:y = - \dfrac{b {x}^{2} }{ {a}^{2} } [/tex]

[tex] \underline{\tt \bold{ \: Answer \: (b)}}[/tex]

We know,

  • [tex]\tt :\longmapsto\:\vec{v} \: = \: \dfrac{d}{dt} \vec{r}[/tex]

and

  • [tex]\tt :\longmapsto\:\vec{w} = \dfrac{d}{dt} \vec{v}[/tex]

As,

  • [tex]\tt :\longmapsto\:\vec{r} = a t \: \hat{i} - b {t}^{2} \: \hat{j}[/tex]

On differentiating with respect to 't', we get

  • [tex]\tt :\longmapsto\:\dfrac{d}{dr} \vec{r} = \dfrac{d}{dt} (a t \: \hat{i} - b {t}^{2} \: \hat{j})[/tex]

  • [tex]\tt :\longmapsto\:\vec{v} = a \: \hat{i} - 2b {t} \: \hat{j}[/tex]

On differentiating with respect to 't', we get

  • [tex]\tt :\longmapsto\:\dfrac{d}{dt} \vec{v} = \dfrac{d}{dt} (a\: \hat{i} - 2b {t} \: \hat{j})[/tex]

  • [tex]\tt :\longmapsto\:\vec{w} = - 2b \: \hat{j}[/tex]

Now,

  • [tex]\tt :\longmapsto\: |\vec{v}| = \sqrt{ {a}^{2} + {( - 2bt)}^{2} } [/tex]

  • [tex]\tt\implies \: |\vec{v}| = \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } [/tex]

And

  • [tex]\tt :\longmapsto\: |\vec{w}| = \sqrt{ {( - 2b)}^{2} } [/tex]

  • [tex]\tt :\longmapsto\: |\vec{w}| = 2b[/tex]

[tex] \underline{\rm \tt{ \: Answer \: (c)}}[/tex]

We know,

To distinguish the angle 'a' with numeric constant 'a',

Let angle between two vectors be 'p'.

Angle between two vector is given by

  • [tex]\tt :\longmapsto\:cosp \: = \dfrac{\vec{v} \: . \: \vec{w}}{ |\vec{v}| \: |\vec{w}| } [/tex]

  • [tex]\tt :\longmapsto\:cosp \: = \dfrac{(a\: \hat{i} - 2bt \: \hat{j}) \: . \: ( - 2b \: \hat{j})}{ \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } \times 2b } [/tex]

  • [tex]\tt :\longmapsto\:cosp = \dfrac{ {4tb}^{2} }{ \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } \times 2b } [/tex]

  • [tex]\tt\implies \:cosp = \dfrac{2bt}{ \sqrt{ {a}^{2} + {4b}^{2} {t}^{2} } } [/tex]

As,

We know

  • [tex]\tt :\longmapsto\: {tan}^{2} p = {sec}^{2} p - 1[/tex]

  • [tex]\tt :\longmapsto\: {tan}^{2} p = \dfrac{ {a}^{2} + 4 {b}^{2} {t}^{2} }{4 {b}^{2} {t}^{2} } - 1[/tex]

  • [tex]\tt :\longmapsto\: {tan}^{2} p = \dfrac{ {a}^{2} + 4 {b}^{2} {t}^{2} - 4 {b}^{2} {t}^{2}}{4 {b}^{2} {t}^{2} } [/tex]

  • [tex]\tt:\longmapsto\: {tan}^{2} p = \dfrac{ {a}^{2}}{4 {b}^{2} {t}^{2} } [/tex]

  • [tex]\tt :\longmapsto\:tanp \: = \: \dfrac{a}{2bt} [/tex]

  • [tex]\tt\implies \:p \: = \: {tan}^{ - 1} \bigg(\dfrac{a}{2bt} \bigg)[/tex]

[tex] \underline{\tt \bold{ \: Answer \: (d)}}[/tex]

The mean velocity vector is given by

  • [tex]\tt :\longmapsto\:\vec{ \overline{v}} = \dfrac{\int_0^t \: \vec{v} \: dt}{ \int \: dt} [/tex]

  • [tex]\tt :\longmapsto\:\vec{ \overline{v}} = \dfrac{\int_0^t \: (a\: \hat{i} - 2bt \: \hat{j}) \: dt}{ t} [/tex]

  • [tex]\tt :\longmapsto\:\vec{ \overline{v}} = \dfrac{at\: \hat{i} - 2b \: \dfrac{ {t}^{2} }{2} \: \hat{j} }{t} [/tex]

  • [tex]\tt :\longmapsto\:\vec{ \overline{v}} = a\: \hat{i} - b {t} \: \hat{j}[/tex]

Hence,

  • [tex]\tt :\longmapsto\: |\vec{ \overline{v}} | = \sqrt{ {a}^{2} + {( - bt)}^{2} } [/tex]

  • [tex]\tt\implies \: |\vec{ \overline{v}}| = \sqrt{ {a}^{2} + {b}^{2} {t}^{2} } [/tex]

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[tex]\qquad\qquad\qquad\qquad\qquad\qquad\tt{fri \: 03-04-2022} \\ \qquad\qquad\qquad\qquad\qquad\qquad\tt{11:44 \: am}[/tex]